blocks|key|7711|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7712|echo+date("Y-m-d",+strtotime('-1+Monday'))." ";
for($i=1;+$i<=13;+$i%2B%2B){
++++echo+date("Y-m-d",+strtotime('%2B'.$i.'+Monday'))." ";
}|code-block|syntax|javascript|7713|或者您只需要一条for语句？|offset|length|style|CODE|7714|entityMap^0|0|0|8|3|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Try this:

<pre><code>echo date("Y-m-d", strtotime('-1 Monday'))."&lt;br&gt;";
for($i=1; $i&lt;=13; $i++){
 echo date("Y-m-d", strtotime('+'.$i.' Monday'))."&lt;br&gt;";
}
</code></pre>

Or you need only one <code>for</code> statement?

blocks|key|7756|text|您可以使用以下命令：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7757|for($i=-1;+$i<=13;+$i%2B%2B){
++++if($i+!==+0){
++++++++echo+date("Y-m-d",+strtotime('%2B'.$i.'+Monday'))." ";++++++++
++++}
}|code-block|syntax|javascript|7758|您只需省略$i为零的情况。|7759|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

You can use this:

<pre><code>for($i=-1; $i&lt;=13; $i++){
 if($i !== 0){
 echo date("Y-m-d", strtotime('+'.$i.' Monday'))."&lt;br&gt;"; 
 }
}
</code></pre>

You just have to omit the case that $i is zero.

blocks|key|7809|text|我会这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7810|for($i+=-+1;+$i+<=+13;+$i+==+0+?+$i+%2B=+2+:+$i%2B%2B){
++++echo+date("Y-m-d",+strtotime("$i+Monday"))+.+" ";
}|code-block|syntax|javascript|7811|使用三元运算符检查$i是否为0-如果是，则将其增加2而不是1+:)|offset|length|style|CODE|7812|entityMap^0|0|0|9|2|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I would do it like this:

<pre><code>for($i =- 1; $i &lt;= 13; $i == 0 ? $i += 2 : $i++){
 echo date("Y-m-d", strtotime("$i Monday")) . "&lt;br&gt;";
}
</code></pre>

Using a ternary operator to check if <code>$i</code> is 0 - and if, increase it by 2 instead of 1 :)

blocks|key|4515295|text|function+mondays()+{
++++$begin+=+new+DateTime('last+monday');
++++$end+=+clone+$begin;
++++$end->add(new+DateInterval('P14W'));+//+next+13+%2B+last

++++$interval+=+new+DateInterval('P1W');
++++$daterange+=+new+DatePeriod($begin,+$interval+,$end);

++++foreach($daterange+as+$date){
++++++++yield+$date;
++++}
}


foreach(mondays()+as+$date){
++++echo+$date->format("Y-m-d"),+PHP_EOL;
}|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|4515296|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>function mondays() {
 $begin = new DateTime('last monday');
 $end = clone $begin;
 $end-&gt;add(new DateInterval('P14W')); // next 13 + last

 $interval = new DateInterval('P1W');
 $daterange = new DatePeriod($begin, $interval ,$end);

 foreach($daterange as $date){
 yield $date;
 }
}


foreach(mondays() as $date){
 echo $date-&gt;format("Y-m-d"), PHP_EOL;
}
</code></pre>

blocks|key|4515386|text|这应该可以工作(你也可以在其他日子测试!=今天)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4515387|//In+order+to+avoid+problems+with+midnight+and+daylight+saving+time
$refTime+=+date("Y-m-d+12:00:00");

for($i=-1;+$i<13;+$i%2B%2B){
++++echo+date("Y-m-d",+strtotime("$refTime+this+Monday+%2B".$i*7+."+days"))." ";
}|code-block|syntax|javascript|4515388|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

This should works (and you can test also other days != today)

<pre><code>//In order to avoid problems with midnight and daylight saving time
$refTime = date("Y-m-d 12:00:00");

for($i=-1; $i&lt;13; $i++){
 echo date("Y-m-d", strtotime("$refTime this Monday +".$i*7 ." days"))."&lt;br&gt;";
}
</code></pre>

blocks|key|7902|text|我的方法如下(通过使用DateTime类)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|7903|$monday+=+new+DateTime('last+monday');+//if+today+is+monday+this+will+return+last+week's
$oneWeek+=+new+DateInterval('P1W');
for($i=0;+$i<=13;+$i%2B%2B){
++++echo+$monday->format('Y-m-d')." ";
++++$monday->add($oneWeek);
}|code-block|syntax|javascript|7904|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

My approach will be the following (by using DateTime class)

<pre><code>$monday = new DateTime('last monday'); //if today is monday this will return last week's
$oneWeek = new DateInterval('P1W');
for($i=0; $i&lt;=13; $i++){
 echo $monday-&gt;format('Y-m-d')."&lt;br&gt;";
 $monday-&gt;add($oneWeek);
}
</code></pre>

I have the following code that return's the next 13 Mondays from today's date.

<pre><code>for($i=1; $i&lt;=13; $i++){
 echo date("Y-m-d", strtotime('+'.$i.' Monday'))."&lt;br&gt;";
}
</code></pre>

I want to be able to amend this so it not only shows the next 13 Mondays but the Monday that has just past.

I tried amending the code as follows but I then get two instances of the next Monday returned.

<pre><code>for($i=-1; $i&lt;=13; $i++){
echo date("Y-m-d", strtotime('+'.$i.' Monday'))."&lt;br&gt;";
}
</code></pre>

Data returned.

<pre><code>2015-04-13
2015-04-20 //&lt;--
2015-04-20
2015-04-27
2015-05-04
2015-05-11
2015-05-18
2015-05-25
2015-06-01
2015-06-08
2015-06-15
2015-06-22
2015-06-29
2015-07-06
2015-07-13
</code></pre>

Any ideas on how I achieve this?

Finding next 13 Mondays and the last Monday

我有下面的代码，返回从今天开始的下13个星期一。for($i=1; $i<=13; $i++){ echo date("Y-m-d", strtotime('+'.$i.' Monday'))." ";}我希望能够修改它，这样它不仅可以显示下一个13个星期一，而且可以显示刚刚过去的星期一。我尝试修改代码，如...

问查找下一个13个星期一和最后一个星期一
EN

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问查找下一个13个星期一和最后一个星期一
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